Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int x[10020];
int y[10020];
vector<pair<int, int> > xop;
vector<pair<int, int> > yop;
void op(int i, int j, int a[], vector<pair<int, int> >& ops) {
  a[i] ^= a[j];
  ops.push_back(make_pair(i, j));
}
int getmsb(int x) {
  int msb = 30;
  while (msb >= 0 && (x & (1 << msb)) == 0) {
    --msb;
  }
  if (msb < 0) return -1;
  return msb;
}
void norm(int a[], vector<pair<int, int> >& ops) {
  int s = 0;
  for (int k = 30; k >= 0; --k) {
    int f = -1;
    for (int i = s; i < n; ++i) {
      if (a[i] & (1 << k)) {
        f = i;
        break;
      }
    }
    if (f == -1) continue;
    if (f != s) {
      op(s, f, a, ops);
    }
    for (int i = s + 1; i < n; ++i) {
      if (a[i] & (1 << k)) {
        op(i, s, a, ops);
      }
    }
    ++s;
  }
  for (int i = s - 1; i >= 0; --i) {
    int msb = getmsb(a[i]);
    if (msb < 0) return;
    for (int j = i - 1; j >= 0; --j) {
      if (a[j] & (1 << msb)) {
        op(j, i, a, ops);
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &y[i]);
  }
  norm(x, xop);
  norm(y, yop);
  int i = 0;
  while (i < n) {
    int xm = getmsb(x[i]);
    int ym = getmsb(y[i]);
    if (xm != ym) {
      if (xm < ym) {
        printf("-1\n");
        return 0;
      }
      for (int j = i - 1; j >= 0; --j) {
        if (y[j] & (1 << xm)) {
          op(j, i, x, xop);
        }
      }
      op(i, i, x, xop);
      int k = i + 1;
      while (k < n && x[k] > 0) {
        op(k - 1, k, x, xop);
        op(k, k, x, xop);
        ++k;
      }
    } else {
      ++i;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (x[i] != y[i]) {
      printf("-1\n");
      return 0;
    }
  }
  int m = xop.size() + yop.size();
  printf("%d\n", m);
  for (int i = 0; i < (int)xop.size(); ++i) {
    printf("%d %d\n", xop[i].first + 1, xop[i].second + 1);
  }
  for (int i = (int)yop.size() - 1; i >= 0; --i) {
    printf("%d %d\n", yop[i].first + 1, yop[i].second + 1);
  }
  return 0;
}